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Mathematics
Mathematics
Problem
Prove that
∑
k
=
1
n
−
1
n
C
k
[
cos
k
x
⋅
cos
(
n
+
k
)
x
+
sin
(
n
−
k
)
x
⋅
sin
(
2
n
−
k
)
x
]
=
(
2
n
−
2
)
cos
n
x
.
\sum_{k=1}^{n-1} {}^n C_k [\cos kx \cdot \cos(n+k)x + \sin(n-k)x \cdot \sin(2n-k)x] = (2^n - 2) \cos nx.
k
=
1
∑
n
−
1
n
C
k
[
cos
k
x
⋅
cos
(
n
+
k
)
x
+
sin
(
n
−
k
)
x
⋅
sin
(
2
n
−
k
)
x
]
=
(
2
n
−
2
)
cos
n
x
.
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